I try to identify conditions for the Fourier-transformation $\mathcal{F}(f)$ of some function $f \in L^1(\mathbb{R}^n)$ to be real-analytic. Namely I want to show that one of the following two conditions is sufficient:

  1. $\exists K \text{ compact}: \text{supp}(f) \subset K$
  2. $\exists C\in (0,\infty): |f|\leq\exp(-C|x|)$

A few notes:

Certainly, what we are looking for is a good bound on the derivatives, in particular $|D^\alpha\mathcal{F}(f)|\leq C_n/r^{|\alpha|}$ where $\alpha$ is a multi-index would be some inequality of the form I am interested in.

I tried to find some resources and found the Paley-Wiener theorems, but I could not find any resource that shows what I am looking for. Rudin shows in "Functional Analysis" (Theorem 7.22+) that if $f$ is smooth and has compact support $\mathcal{F}(f)$ is holomorphic and the restriction of it to the reals is real-analytic then. However, I want to find a proof that doesn't rely on complex-analysis and I would need to remove the restriction that $f$ is smooth.

I currently work with the following definition of the Fourier-transformation:

$$\mathcal{F}(f)(t) = \int_{\mathbb{R}^n} f(x)\, e^{-2\pi \mathrm{i} t \cdot x} \,\mathrm{d} x.$$

  • $\begingroup$ In the two case you can compute the derivatives of the Fourier transform thanks to the dominated convergence theorem. $\endgroup$ – Davide Giraudo Apr 24 '12 at 19:06
  • $\begingroup$ @DavideGiraudo, Thank you for the hint, I know that I can pull the derivate into the integral, but the problem is that $f$ is not differentiable therefore I do not know how to go on from there. $\endgroup$ – Listing Apr 24 '12 at 19:14
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    $\begingroup$ You differentiate with respect to $t$, to the differentiation is on the exponential. $\endgroup$ – Davide Giraudo Apr 24 '12 at 19:15
  • $\begingroup$ @DavideGiraudo Of course thank you, I will make to sure to try that in detail later when I have some time. $\endgroup$ – Listing Apr 24 '12 at 19:29
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    $\begingroup$ By the way, Condition 2 implies that the Fourier transform is the restriction to $\mathbb R$ of a function analytic in the strip $|Im(z)|<B$ for all $B<C$. $\endgroup$ – B R Apr 24 '12 at 19:52

In the two cases, we can apply the dominated convergence theorem to show by induction on $|\alpha|$ that $$\partial^{\alpha}\mathcal F(f)(t)=\int_{\mathbb R^n}f(x)(-2\pi i)^{\alpha}x^{\alpha}e^{-2\pi it\cdot x}dx,$$ hence $$|\partial^{\alpha}\mathcal F(f)(t)|=(2\pi)^{\alpha}\int_{\mathbb R^n}|f(x)|\cdot |x|^{\alpha}dx.$$ The last function is integrable in the two cases, so the last inequality ensures us that under 1. or 2. the Fourier transform of $f$ is real-analytic.

  • $\begingroup$ Thank you, the solution is more straight forward than I thought. $\endgroup$ – Listing Apr 24 '12 at 19:58
  • $\begingroup$ @davide giraudo, This is establishes $\mathscr{F}(f)$ is smooth, but where does the analytic part come from? Is it that trivial? $\endgroup$ – dezdichado Apr 20 '16 at 4:30

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